Hint:
Start by using the power reduction formula to get rid of the sine-squared term:
$$\begin{align}
\mathcal{I}_{n}
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{nx}}{\sin{x}}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{2nx}}{2\sin{x}}\,\mathrm{d}x.\\
\end{align}$$
Then, writing the $\cos{2nx}$ as a binomial series, we can rewrite the integrand as a polynomial of sines and cosines:
$$\cos{(2nx)}=\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=1-\sum_{k=0}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=1-\cos^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies 1-\cos{(2nx)}=\sin^2{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k}{(x)}$$
$$\implies \frac{1-\cos{(2nx)}}{\sin{(x)}}=\sin{(x)}-\sum_{k=1}^{n}(-1)^{k}\binom{2n}{2k}\cos^{2(n-k)}{(x)}\sin^{2k-1}{(x)}.$$
Finally, integrate term-by-term.